\newproblem{lay:7_4_24}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 7.4.24}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Using the notation of Exercise 7.4.23, show that $A^T\mathbf{u}_j=\sigma_j\mathbf{v}_j$.
}{
   % Solution
	Let's calculate first $A^T$
	\begin{center}
		$A^T=(\sigma_1\mathbf{u}_1\mathbf{v}_1^T+\sigma_2\mathbf{u}_2\mathbf{v}_2^T+...+\sigma_r\mathbf{u}_r\mathbf{v}_r^T)^T=
		   \sigma_1\mathbf{v}_1\mathbf{u}_1^T+\sigma_2\mathbf{v}_2\mathbf{u}_2^T+...+\sigma_r\mathbf{v}_r\mathbf{u}_r^T$
	\end{center}
	Now, we can easily calculate $A^T\mathbf{u}_j$
	\begin{center}
		$A^T\mathbf{u}_j=\left(\sum\limits_{i=1}^r{\sigma_i\mathbf{v}_i\mathbf{u}_i^T}\right)\mathbf{u}_j=
		   \sum\limits_{i=1}^r{\sigma_i\mathbf{v}_i\mathbf{u}_i^T\mathbf{u}_j}$
	\end{center}
	Since the columns of $U$ are orthogonal to each other all products $\mathbf{u}_i^T\mathbf{u}_j$ are 0 if $i\neq j$ and 1 if $i=j$. Then, the previous sum reduces to
	\begin{center}
		$A^T\mathbf{u}_j=\sigma_j\mathbf{v}_j$
	\end{center}
}
\useproblem{lay:7_4_24}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}

